# Find all integer solutions for $x^3+1=y^2$. [duplicate]

Find all integer solutions for $$x^3+1=y^2.$$

Attempt: By guessing, I found five pairs of integer solutions for the equation: $$(2, \pm 3)$$, $$(0, 1)$$, $$(-1, 0)$$ and $$(0, -1)$$, but really I don't know how to solve it analytically without guessing. Some people lead this problem to the Catalan?

Hint: This an example of "Mordell's Equation" - curves of the form $$y^2 = x^3 + D$$ (in your case D = 1). Many things are known about its integral solutions. You might find this article useful , you can also proceed using this answer and also this one

Addendum: Mordell spent many years of his life studying integral solutions of the equation $$y^2 = x^3 + k$$, where $$k$$ is a fixed nonzero integer. The equation could be justified as having interest because it's one of the simplest examples of an elliptic curve, but it's important for a better reason. The $$abc$$-conjecture, which has connections to many other problems, does not at first look like it is about Mordell's equation. However, the $$abc$$ conjecture turns out to be equivalent to specific upper bounds on relatively prime integral solutions $$(x,y)$$ to Mordell's equation $$y^2 = x^3 + k$$ in terms of the parameter $$k$$. So, as Barry Mazur once remarked, the Mordell equation is a far more central topic to all of number theory than its rather special appearance suggests.

• This is not a hint, this is a reference. – Wojowu Jun 2 at 22:56
• The titled question should be follow the linked answer we do not need to repeat the same steps which are montioned in the linked answer – zeraoulia rafik Jun 2 at 22:57
• I am not complaining about the answer per se (though I could, this is something I would see more fit as a comment). I am only complaining about the use of the word "Hint" to refer to something that's a link to a full solution and not a hint. – Wojowu Jun 2 at 23:06

The elementary approach is to write

$$x^3=y^2-1=(y-1)(y+1)$$

Now, when $$y$$ is even then $$y-1$$ and $$y+1$$ are relatively prime, so $$y-1=w^3$$ and $$y+1=z^3$$ for integers $$w,z.$$ But $$z^3-w^3=2$$ is only possible if $$(w,z)=(-1,1),$$ or $$y=0, x=-1.$$

The case $$y$$ odd is a bit harder, but not too hard. It becomes equivalent to solving the equation: $$2n^3=m(m+1)$$ where $$n=x/2$$ and $$m=(y-1)/2.$$

Use that $$m$$ and $$m+1$$ are relatively prime.

This reduces to the case $$u^3-2v^3=\pm1.\tag{1}$$ That’s a harder equation to solve, perhaps.

Once you have $$u,v$$, you get $$y=u^3+2v^3$$ and $$x=2uv.$$

The solutions to (1) are $$(u,v)=(1,1),(-1,-1),(1,0),(-1,0)$$ yield $$(x,y)=(2,3),(2,-3),(0,1),$$ and $$(0,-1)$$ respectively.

That there are no other solutions seems non-trivial. Perhaps you can rewrite it as:

$$(x-y)(x^2+xy+y^2)=x^3-y^3=y^3\pm 1=(y\pm 1)(y^2\mp y+1)$$

but I'm not seeing an obvious approach from there.

According to Sage, all integral points are below.

E=EllipticCurve([0,0,0,0,1])
E.integral_points()

$$(0, \pm 1),(2, \pm 3),(-1,0)$$